JEE Main 2021
LEVELJEE Main
Visualized Solution
The Sigma Insight: Plane Mirror
The problem of finding the field of view in a plane mirror is a classic application of geometric optics. It beautifully combines the physics of reflection with the mathematics of similar triangles. Let's break down the solution step-by-step.
Analyzing the Setup
Imagine you are standing in a room with a plane mirror of width hanging on the wall. A point source of light, , is placed exactly in front of the mirror's center. You are walking along a straight path parallel to the mirror, at a distance of (or ) from it.
Our goal is to find the exact stretch of distance along your path where you can see the reflection of the light source in the mirror.
The Master Equation
Field of View
The field of view is strictly bounded by the extreme rays of light. These are the rays that travel from the source , hit the very edges of the mirror (let's call them and ), and reflect towards your path.
Let the reflected rays hit your path at points and . You will only be able to see the image of the source when you are standing anywhere between and .
To find this distance , we use a clever geometric trick involving the virtual image. We know that for a plane mirror, the virtual image is formed at the exact same distance behind the mirror as the object is in front of it. Since the source is in front, the virtual image is formed behind the mirror.
Applying Similar Triangles
When we extend the reflected rays backwards, they all appear to intersect at the virtual image . This creates two perfectly similar triangles:
1. The smaller triangle , formed by the image and the mirror.
2. The larger triangle , formed by the image and the extreme points on your path.
Because these triangles are similar (), the ratio of their bases must equal the ratio of their heights.
Let's identify these dimensions:
- The base of the small triangle is the mirror's width, .
- The height of the small triangle is the image distance, .
- The base of the large triangle is our unknown distance, .
- The height of the large triangle is the total distance from the image to your path. This is the image distance () plus the distance from the mirror to your path (), giving a total height of .
Final Calculation
Now, we simply set up the ratio:
Substituting our known values into the equation:
The right side of the equation simplifies beautifully:
Finally, multiplying both sides by , we get:
The distance between the extreme points where the man can see the image is . This elegant geometric approach saves us from complex trigonometric calculations and gives us a clean, intuitive result!
Similar Questions
LEVELJEE Main
A point source of light , placed at a distance in front of the centre of a plane mirror of width , hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance from it as shown. The greatest distance over which he can see the image of the light source in the mirror is
(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Main
A point source of light, is placed at a distance in front of the centre of plane mirror of width which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance as shown below. The distance over which the man can see the image of the light source in the mirror is
(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main
Two plane mirrors and are at right angle to each other shown. A point source is placed at and meter away from and , respectively. The shortest distance between the images thus formed is (Take )
(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Advanced
The plane mirrors ( and ) are inclined to each other such that a ray of light incident on mirror and parallel to the mirror is reflected from mirror parallel to the mirror . The angle between the two mirror is
(A)
(B)
(C)
(D)
JEE Advanced 2013
LEVELJEE Main
A ray of light travelling in the direction is incident on a plane mirror. After reflection, it travels along the direction . The angle of incidence is
(A)
30^\circ
(B)
45^\circ
(C)
60^\circ
(D)
75^\circ
JEE Advanced 2022
LEVELJEE Advanced
Three plane mirrors form an equilateral triangle with each side of length L. There is a small hole at a distance from one of the corners as shown in the figure. A ray of light is passed through the hole at an angle and can only come out through the same hole. The cross section of the mirror configuration and the ray of light lie on the same plane.
* Multiple Correct Options
(A)
The ray of light will come out for , for .
(B)
There is an angle for at which the ray of light will come out after two reflections.
(C)
The ray of light will NEVER come out for and .
(D)
The ray of light will come out for , and after six reflections.
LEVELJEE Main
Two plane mirrors and are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle at a point just inside one end of . The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
(A)
28
(B)
30
(C)
32
(D)
34
LEVELJEE Main
If two mirrors are kept at to each other, then the number of images formed by them is
(A)
5
(B)
6
(C)
7
(D)
8
LEVELBoard
To get three images of a single object, one should have two plane mirrors at an angle of
(A)
(B)
(C)
(D)